I have a tensor of size m x n (m rows and n columns).
For example:
[ 5 8 4 3
1 3 5 4
3 9 8 6 ]
I wish to randomly select half of the columns, and set all the values in this columns as zeros.
For our example, it will create something like this:
[ 5 0 4 0
1 0 5 0
3 0 8 0 ]
I'm aware how to set zero randomly half of all the elements,
torch.rand(x.shape) > 0.5
but done randomly without consideration in the columns, which is not helpfull for my case.
Thank you for any help,
Dave
import torch
x = torch.rand(3,4)
x
tensor([[0.0143, 0.1070, 0.9985, 0.0727],
[0.4052, 0.8716, 0.7376, 0.5495],
[0.2553, 0.2330, 0.9285, 0.6535]])
for i in [1,3] : # list has your columns which you want to make zero
x[:,i] = 0
Related
Having df of probabilities distribution, I get max probability for rows with df.idxmax(axis=1) like this:
df['1k-th'] = df.idxmax(axis=1)
and get the following result:
(scroll the tables to the right if you can not see all the columns)
0 1 2 3 4 5 6 1k-th
0 0.114869 0.020708 0.025587 0.028741 0.031257 0.031619 0.747219 6
1 0.020206 0.012710 0.010341 0.012196 0.812495 0.113863 0.018190 4
2 0.023585 0.735475 0.091795 0.021683 0.027581 0.054217 0.045664 1
3 0.009834 0.009175 0.013165 0.016014 0.015507 0.899115 0.037190 5
4 0.023357 0.736059 0.088721 0.021626 0.027341 0.056289 0.046607 1
the question is how to get the 2-th, 3th, etc probabilities, so that I get the following result?:
0 1 2 3 4 5 6 1k-th 2-th
0 0.114869 0.020708 0.025587 0.028741 0.031257 0.031619 0.747219 6 0
1 0.020206 0.012710 0.010341 0.012196 0.812495 0.113863 0.018190 4 3
2 0.023585 0.735475 0.091795 0.021683 0.027581 0.054217 0.045664 1 4
3 0.009834 0.009175 0.013165 0.016014 0.015507 0.899115 0.037190 5 4
4 0.023357 0.736059 0.088721 0.021626 0.027341 0.056289 0.046607 1 2
Thank you!
My own solution is not the prettiest, but does it's job and works fast:
for i in range(7):
p[f'{i}k'] = p[[0,1,2,3,4,5,6]].idxmax(axis=1)
p[f'{i}k_v'] = p[[0,1,2,3,4,5,6]].max(axis=1)
for x in range(7):
p[x] = np.where(p[x]==p[f'{i}k_v'], np.nan, p[x])
The loop does:
finds the largest value and it's column index
drops the found value (sets to nan)
again
finds the 2nd largest value
drops the found value
etc ...
in my data frame I want to iterrows() of two columns but want to save result in 1 column.for example df is
x y
5 10
30 445
70 32
expected output is
points sequence
5 1
10 2
30 1
445 2
I know about iterrows() but it saved out put in two different columns.How can I get expected output and is there any way to generate sequence number according to condition? any help will be appreciated.
First never use iterrows, because really slow.
If want 1, 2 sequence by number of columns convert values to numy array by DataFrame.to_numpy and add numpy.ravel, then for sequence use numpy.tile:
df = pd.DataFrame({'points': df.to_numpy().ravel(),
'sequence': np.tile([1,2], len(df))})
print (df)
points sequence
0 5 1
1 10 2
2 30 1
3 445 2
4 70 1
5 32 2
Do this way:
>>> pd.DataFrame([i[1] for i in df.iterrows()])
points sequence
0 5 1
1 10 2
2 30 1
3 445 2
I am looking for a pythonic way of replacing values based on whether values are big of small. Say I have a data frame:
ds = pandas.DataFrame({'x' : [4,3,2,1,5], 'y' : [4,5,6,7,8]})
I'd like to replace values on x which are lower than 2 by 2 and values higher than 4 by 4. And similarly with y values, replacing values lower than 5 by 5 and values higher than 7 by 7 so as to get this data frame:
ds = pandas.DataFrame({'x' : [4,3,2,2,4], 'y' : [5,5,6,7,7]})
I did it by iterating on the rows but is really ugly, any more pandas-pythonic way (Basically I want to eliminate extreme values)
You can check with clip
ds.x.clip(2,4)
Out[42]:
0 4
1 3
2 2
3 2
4 4
Name: x, dtype: int64
#ds.x=ds.x.clip(2,4)
#ds.y=ds.y.clip(5,7)
One way of doing this as follows:
>>> ds[ds.x.le(2) ] =2
>>> ds[ds.x.ge(4) ] =4
>>> ds
x y
0 4 4
1 3 5
2 2 6
3 2 2
4 4 4
If I have a dataframe of the form:
tag element_id
1 12
1 13
1 15
2 12
2 13
2 19
3 12
3 15
3 22
how can I compute the overlaps of the tags in terms of the element_id ? The result I guess should be an overlap matrix of the form:
1 2 3
1 X 2 2
2 2 X 1
3 2 1 X
where I put X on the diagonal since the overlap of a tag with itself is not relevant and where the numbers in the matrix represent the total element_ids that the two tags share.
My attempts:
You can try and use a for loop like :
for item in df.itertuples():
element_lst += [item.element_id]
element_tag = item.tag
# then intersect the element_list row by row.
# This is extremely costly for large datasets
The second thing I was thinking about was to use df.groupby('tag') and try to somehow intersect on element_id, but it is not clear to me how I can do that with grouped data.
merge + crosstab
# Find element overlap, remove same tag matches
res = df.merge(df, on='element_id').query('tag_x != tag_y')
pd.crosstab(res.tag_x, res.tag_y)
Output:
tag_y 1 2 3
tag_x
1 0 2 2
2 2 0 1
3 2 1 0
I have a super strange problem which I spent the last hour trying to solve, but with no success. It is even more strange since I can't replicate it on a small scale.
I have a large DataFrame (150,000 entries). I took out a subset of it and did some manipulation. the subset was saved as a different variable, x.
x is smaller than the df, but its index is in the same range as the df. I'm now trying to assign x back to the DataFrame replacing values in the same column:
rep_Callers['true_vpID'] = x.true_vpID
This inserts all the different values in x to the right place in df, but instead of keeping the df.true_vpID values that are not in x, it is filling them with NaNs. So I tried a different approach:
df.ix[x.index,'true_vpID'] = x.true_vpID
But instead of filling x values in the right place in df, the df.true_vpID gets filled with the first value of x and only it! I changed the first value of x several times to make sure this is indeed what is happening, and it is. I tried to replicate it on a small scale but it didn't work:
df = DataFrame({'a':ones(5),'b':range(5)})
a b
0 1 0
1 1 1
2 1 2
3 1 3
4 1 4
z =Series([random() for i in range(5)],index = range(5))
0 0.812561
1 0.862109
2 0.031268
3 0.575634
4 0.760752
df.ix[z.index[[1,3]],'b'] = z[[1,3]]
a b
0 1 0.000000
1 1 0.812561
2 1 2.000000
3 1 0.575634
4 1 4.000000
5 1 5.000000
I really tried it all, need some new suggestions...
Try using df.update(updated_df_or_series)
Also using a simple example, you can modify a DataFrame by doing an index query and modifying the resulting object.
df_1
a b
0 1 0
1 1 1
2 1 2
3 1 3
4 1 4
df_2 = df_1.ix[3:5]
df_2.b = df_2.b + 2
df_2
a b
3 1 5
4 1 6
df_1
a b
0 1 0
1 1 1
2 1 2
3 1 5
4 1 6